MCAT Biology : Organic Chemistry, Biochemistry, and Metabolism

Study concepts, example questions & explanations for MCAT Biology

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Example Questions

Example Question #21 :Organic Analyses And Lab Techniques

Chromatography involves the separation of a mixture by allowing a mobile phase to travel along a stationary phase. In thin layer chromatography (TLC), a liquid solution is able to travel along a stationary plate. The distance that a particular compound travels compared to another compound can be determined by comparing the Rffactors for each compound. The Rffactor is determined by dividing the compound's distance by the total distance of the solvent.

Another type of chromatography commonly used is gas chromatography. How is this type of chromatography different from thin layer chromatography?

Possible Answers:

The stationary and mobile phases are in different phases

In gas chromatography, nonpolar compounds move a shorter distance than polar compounds

Gas chromatography takes advantage of the compounds' different melting points

Only thin layer chromatography can determine the purity of a solution

Correct answer:

The stationary and mobile phases are in different phases

Explanation:

Gas chromatography and thin layer chromatography differ by the phases of both the mobile and stationary phases. In gas chromatography, the mobile phase is a gas and the stationary phase is a liquid. In thin layer chromatography, the mobile phase is a liquid and the stationary phase is a solid.

Example Question #1 :Chromatography

A chemist carries out the synthetic scheme shown below. Unfortunately, the first two reactions are incomplete, and a mixture of compounds A, B, and C is obtained after the second step. The chemist purifies this mixture by normal phase chromatography, using silica gel as a stationary phase and a 10:1 hexanes-diethyl ether (v:v) solution as an eluent. In what order would compounds A, B, and C elute off the column?

For each choice, the first compound to elute is listed first.

Mcat_1

Possible Answers:

B, C, A

C, B, A

C, A, B

B, A, C

A, C, B

Correct answer:

C, A, B

Explanation:

In the normal phase chromatography system described, the most nonpolar compound would elute first and the most polar compound would elute last. The silica stationary phase will interact with more polar molecules, while the hexane mobile phase will carry nonpolar molecules. This would slow the progress of polar molecules as they bond to the silica, and enhance the progress of nonpolar molecules as they interact with the mobile phase.

Compound C is the most nonpolar compound because it contains only hydrogen and carbon. Compounds A and B are more polar because of the presence of oxygen, and hence the presence of polarized carbon-oxygen bonds. The alcohol group of compound B makes this compound the most polar of the three molecules by virtue of hydrogen bonding capabilities as well as the carbon-oxygen dipole. Compound B would thus elute last.

Example Question #22 :Organic Analyses And Lab Techniques

Chromatography involves the separation of a mixture by allowing a mobile phase to travel along a stationary phase. In thin layer chromatography (TLC), a liquid solution is able to travel along a stationary plate. The distance that a particular compound travels compared to another compound can be determined by comparing the Rffactors for each compound. The Rffactor is determined by dividing the compound's distance by the total distance of the solvent.

Consider a mystery compound with an Rffactor of 0.6. Which of the following statements would you predict to be accurate?

Possible Answers:

A more nonpolar compound can travel twice the distance of the mystery compound

A more polar compound would have a lower Rffactor than the mystery compound

A more nonpolar compound can have twice the Rffactor of the mystery compound

A compound with a higher molecular weight will travel a farther distance than the mystery compound

Correct answer:

A more polar compound would have a lower Rffactor than the mystery compound

Explanation:

The Rffactor is a ratio of the compound's travel distance compared to the solvent's travel distance; this value will always be between 0 and 1. As a result, a nonpolar compound would not be able to travel twice the distance or have twice the Rffactor as the mystery compound. This would result in a value of 1.2.

The only thing we can predict confidently is that a more polar compound would travel a shorter distance than the mystery compound, which result in a lower Rffactor.

Example Question #1 :Reactions With Nitrogen Compounds

Which of the following statements is true concerning the Hofmann elimination reaction?

Possible Answers:

The Zaitsev product is favored in the elimination reaction

The ammonium is eliminated following an E1 mechanism

The least substituted alkene is the major product in the reaction

The quaternary ammonium salt is a poor leaving group

Correct answer:

The least substituted alkene is the major product in the reaction

Explanation:

The Hofmann product is the most favored product in a Hofmann elimination reaction. The reaction follows an E2 mechanism, where a quaternary ammonium salt is able to be removed from a hydrocarbon, resulting in an alkene product. This reaction results in the least substituted alkene being the primary product.

Example Question #1 :Reactions With Ketones And Aldehydes

Which of the following statements is false concerning the Wittig reaction?

Possible Answers:

The ketone or aldehyde is bound by the negatively charged phosphorous atom

的刺激uct compound that includes the phosphorous from the ylide will be attached to the reactant's carbonyl oxygen

The ylide attaches to the substrate via nucleophilic addition

A mixture of cis and trans isomers are created in the reaction

Correct answer:

The ketone or aldehyde is bound by the negatively charged phosphorous atom

Explanation:

The Wittig reaction is used to make an alkene from a ketone or aldehyde. The ylide (RPPh3) attaches to the substrate via nucleophilic addition, with the negative carbanion attaching to the ketone or aldehyde. The phosphorus atom carries a positive charge, balancing the carbanion.

其他所有列出的答案是正确的选择。的刺激ucts of the reaction are an alkene (cis and trans) with the addition of the R group of the ylide and Ph3PO.

Example Question #1 :Help With Wittig Reactions

All of the following are characteristics of a Wittig reaction except__________.

Possible Answers:

it produces a trialkylphosphine oxide or triarylphosphine oxide as a by-product

it involves the reaction of a phosphonium ylide with a carbonyl

it proceeds through a phosphaoxetane intermediate

it results in the formation of a carbon-carbon double bond

it results in the exclusive formation oftransdouble bonds

Correct answer:

it results in the exclusive formation oftransdouble bonds

Explanation:

The Wittig reaction involves the reaction of a phosphonium ylide (generated by treating a phosphonium salt with a strong base) with a ketone or aldehyde.

The reaction proceeds through a phosphaoxetane (4-membered ring containing both phosphorus and oxygen) intermediate to generate a new compound containing a carbon-carbon double bond, plus a phosphine oxide byproduct. It does not formtransdouble bonds exclusively; sometimes, a mixture ofcisandtransisomers are obtained, and sometimes thecisisomer is the predominant product.

Example Question #1 :Nucleophilic Addition

What is created when a ketone is reacted with a phosphorus ylide?

Possible Answers:

Aldehyde

Alkene

Ester

Alkane

Correct answer:

Alkene

Explanation:

The Wittig reaction involves a ketone or aldehyde reacting with a phosphorus ylide, a molecule with a negatively charged carbanion. The ketone will undergo nucleophilic addition and form a betaine. This intermediate will then form an alkene with a triphenylphosphine oxide being released. The Wittig reaction will form a mixture of both cis and trans isomers if the carbanion has two different substituents.

Wittig general reaction:

Example Question #3 :Reactions With Ketones And Aldehydes

The reaction between one mole of acetone and one mole of 1-propanol in aqueous acid will result in the formation of what product?

Possible Answers:

A ketal compound

An acetal compound

A hemiketal compound

A hemiacetal compound

Correct answer:

A hemiketal compound

Explanation:

A hemiketal compound is the result of nucleophilic attack by an alcohol (1-propanol) on the carbonyl of a ketone (acetone). The previously double-bonded oxygen now bears a negative charge, and deprotonates the now positively charged attacking alcohol. The hemiketal of acetone will have a hydroxy group and an -OCH3group bound to the central carbon.

Example Question #4 :Reactions With Ketones And Aldehydes

Imine

What is the product of the above reaction?

Possible Answers:

An enamine

An imine

No reaction occurs

An amide compound

An amino compound

Correct answer:

An imine

Explanation:

This reaction shows nucleophilic acyl attack on a carbonyl group by the ammonia molecule. This reaction leads to the loss of oxygen as water, and eventually the formation of an imine. Note the difference between each product type. An imine is a nitrogen with one nitrogen-carbon double bond and one substituent. An enamine is formed from a nitrogen with no double bonds. An amide is a nitrogen bound to a carbonyl carbon.

Imine_reaction

Example Question #5 :Reactions With Ketones And Aldehydes

An aldehyde undergoing one round of nucleophilic attack by an equivalent of alcohol results in a(n)__________product, where two rounds of nucleophilic attack (two alcohol equivalents) results in a(n)__________product.

Possible Answers:

hemiketal . . . ketal

hemiacetal . . . acetal

acetal . . . hemiacetal

acetal . . . ketal

Correct answer:

hemiacetal . . . acetal

Explanation:

The correct answer is hemiacetal and acetal. The partial positive character of a carbonyl carbon makes it susceptible to nucleophilic attack. An alcohol's oxygen has free electrons, and therefore can serve as a nucleophile. As the bond between the attacking alcohol's oxygen and the carbonyl carbon forms, the pi bond of the carbonyl group is lost. Once the attacking alcohol's oxygen is deprotonated, the resulting product is an ether (from the attacking alcohol) and a new hydroxyl group (from the now protonated carbonyl oxygen). The resulting molecule is referred to as the hemiacetal product.

Subsequent addition of alcohol leads to the loss of the hydoxyl group as water, resulting in the second ether chain to the original carbon, which is referred to as the acetal product.

The same reaction takes place with ketones as well. With ketones, the first addition of alcohol results in the hemiketal product, and the second addition of alcohol results in the ketal product.

It is important to remember the following material for analyzing acetal and ketal reactions.

1. These alcohol addition reactions only occur with aldehydes and ketones.

2. Hemiacetal and hemiketal products can be identified by having one ether chain and one hydroxyl group attached to the original carbonyl carbon. Acetal and ketal groups have two ether groups attached to the original carbonyl carbon.

3. Acetal and hemiacetal groups will have a hydrogen attached to the original carbonyl carbon, where ketal and hemiketal groups will have carbon chains attached to the original carbonyl carbon (with no bound hydrogens).

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