MCAT Biology : Organic Chemistry, Biochemistry, and Metabolism

Study concepts, example questions & explanations for MCAT Biology

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Example Questions

Example Question #6 :Reactions With Ketones And Aldehydes

Which of the compounds below couldnotbe made from an aldehyde reduction?

Aldehyde_reduction

Possible Answers:

III only

I, III, and IV

I only

II only

IV only

Correct answer:

II only

Explanation:

Of the choices given, all can be made from some type of aldehyde reduction except choice II. Choices I, III, and IV each have the terminal (or primary) alcohol that is characteristic of a former aldehyde. In contrast, choice II has a secondary alcohol, characteristic of a former ketone. In other words, if choice II was oxidized then the product would be a ketone, not an aldehyde.

Example Question #1 :Help With Wolff Krishner Reduction

Acetaldehydeundergoes a Wolf-Kishner reaction, which is the addition of hydrazinewith subsequent addition of a base and heat. In this reaction, the aldehyde is__________, resulting in a(n)__________product.

Possible Answers:

oxidized . . . carboxylic acid

reduced . . . alkane

oxidized . . . amide

reduced . . . alcohol

Correct answer:

reduced . . . alkane

Explanation:

The correct answer is that the aldehyde is reduced to an alkane. In viewing the final product, we see that acetaldehyde would be reduced to ethane. The reaction of any aldehyde or ketone with hydrazine and the subsequent addition of base and heat will result in that aldehyde or ketone being reduced to an alkane, and is referred to as the Wolf-Kishner reaction. The Wolf-Kishner reagent is a commonly tested reducing agent.

Example Question #1 :Reactions With Carboxylic Acids

CompoundBis dissolved in methylene chloride, and then treated with trifluoroacetic acid. Over the next thirty minutes, gas evolution was observed from the reaction mixture. What gas was being given off?

Mcat__4

Possible Answers:


Correct answer:


Explanation:

Treatment of a carboxylic acid with acid results in decarboxylation, and the evolution of, especially if the resulting compound contains a benzylic or allylic carbon, as is the case here.

Example Question #1 :Reactions With Hydrocarbons

The most stable radical shown below is__________. The least stable radical shown below is__________.

Mcat_6

Possible Answers:

radical C . . . radical E

radical D . . . radical A

radical D . . . radical E

radical B . . . radical E

radical E . . . radical C

Correct answer:

radical C . . . radical E

Explanation:

Radical stability increases as carbon substitution increases. In addition, radicals in conjugation with double bonds via resonance are more stable than the corresponding non-conjugated radical.

In this case radicals B, C, and D are all tertiary radicals, but only radical C has additional stabilization from resonance. Radical C is therefore the most stable. Radical E is the least substituted of the five radicals, and is the least stable.

Example Question #2 :Reactions With Hydrocarbons

What intermediate is involved in the conversion of compoundBto compoundC?

Mcat_1

Possible Answers:

Secondary radical

Tertiary carbanion

Secondary carbocation

Tertiary radical

Tertiary carbocation

Correct answer:

Tertiary carbocation

Explanation:

The strong sulfuric acid protonates the hydroxyl group of compound B, resulting in the loss of water as a leaving group and the generation of a carbocation intermediate. Since this carbocation carbon is attached to three other carbons, this is a tertiary carbocation. It is bound to the phenyl substituent, a methyl group, and the branched carbon chain.

Example Question #3 :Reactions With Hydrocarbons

The combustion of pentane with oxygen gas is an exothermic reaction that produces carbon dioxide and water as products. Is this a spontaneous reaction?

Possible Answers:

Yes, only at high temperatures

Yes, only at low temperatures

No, never

Yes, at all temperatures

Correct answer:

Yes, at all temperatures

Explanation:

A balanced equation of the combustion of pentane indicates that one mole of pentane reacts with eight moles of oxygen gas to produce five moles of carbon dioxide and six moles of water.

因为有九个反应物的物质的量和过他n moles of product, entropy increases in this reaction. Exothermic reactions that increase entropy are favorable at all temperatures, as seen in the Gibb's free energy equation.

In our scenario, H is negative and S is positive.

Example Question #4 :Reactions With Hydrocarbons

Which of the following statements concerning the combustion of cycloalkanes is false?

Possible Answers:

As the number of carbons in the ring increases, the molar heat of combustion increases

Oxygen is necessary in order for a combustion reaction to take place

The molar heat of combustion for cyclooctane is nearly double the molar heat of combustion for cyclobutane

The heat of combustion perin cyclohexane is greater than the heat of combustion perin cyclobutane

Correct answer:

The heat of combustion perin cyclohexane is greater than the heat of combustion perin cyclobutane

Explanation:

The heat of combustion for cycloalkanes can be quantified based on two factors: the number of carbons in the ring, and the amount of strain in the ring. Cyclohexane does in fact have a larger molar heat of combustion than cyclobutane. This is because there are more carbons in the ring; however, ring stability will determine the heat of combustion pergroup in the ring. Even though cyclohexane has more carbons than cyclobutane, the heat of combustion pergroup in cyclobutane will be greater compared to cyclohexane due to cyclobutane's ring strain.

Example Question #5 :Reactions With Hydrocarbons

Benzene

Benzene

1_3_5_hex-tri-ene

1,3,5-hexatriene

Which of the compounds shown above will have a greater heat of reduction (hydrogenation)?

Possible Answers:

More information is required

Benzene

The compounds will have the same heat of reduction

1,3,5-hexatriene

Correct answer:

1,3,5-hexatriene

Explanation:

The answer is 1,3,5-hexatriene. Chemicals that are more stable will give off less heat when they are reduced or hydrogenated. Although these two compounds contain the same number of pi bonds to be reduced, benzene is aromatic, and therefore is much more stable than the conjugated non-cyclic hexatriene. We would find that the heat of hydrogenation for hexatriene would be noticeably greater than that of benzene.

Example Question #6 :Reactions With Hydrocarbons

Alkenes A through D contain only carbon and hydrogen. They have the following heats of combustion at.

Alkene A:

Alkene B:

Alkene C:

Alkene D:

排名这四种烯烃的怪兽ms of their stability, with the least stable compound on the left, and the most stable compound on the right.

Possible Answers:

D < B < C < A

A < B < C < D

D < C < B < A

A < C < B < D

A < D < B < C

Correct answer:

D < B < C < A

Explanation:

A combustion reaction of any hydrocarbon yields the same products: carbon dioxide and water. The heat of combustion for the reaction shows how much energy is released as the hydrocarbon is converted to those products.

因为每个烯烃产品是一样的,ny difference in heat of combustion will arise from differences in energy of the starting alkenes. Since the heat of combustion is negative, that means the reactants are higher in energy than the products, creating an exothermic reaction. Furthermore, the more negative the heat of combustion, the higher in energy the reactants are, and the less stable they are as well. Since compound D has the largest heat of combustion, it is highest in energy and therefore the least stable. By similar reasoning, B is the next highest in energy, followed by C, and then finally A, the most stable compound.

Example Question #1 :Reactions With Hydrocarbons

Which of the following steps of free radical chlorination does not produce a free radical as a product?

Possible Answers:

Halogenation

Propagation

Termination

Initiation

Correct answer:

Termination

Explanation:

The three steps of a free radical chlorination reaction are, in order, initiation, propagation, and termination.

Free radicals are produced in the initiation and propagation steps. The termination steps combine any two free radicals formed in the reaction to produce a compound that has no unpaired electrons (free radicals).

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