MCAT Biology : Organic Chemistry, Biochemistry, and Metabolism

Study concepts, example questions & explanations for MCAT Biology

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Example Questions

Example Question #1 : Reactions

Which of the compounds below could not be made from an aldehyde reduction?

Aldehyde_reduction

Possible Answers:

I only

III only

IV only

II only

I, III, and IV

 

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 : Reactions With Ketones And Aldehydes

Acetaldehyde  undergoes a Wolf-Kishner reaction, which is the addition of hydrazine  with subsequent addition of a base and heat. In this reaction, the aldehyde is __________, resulting in a(n) __________ product.

Possible Answers:

reduced . . . alkane

oxidized . . . carboxylic acid

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 #3 : Reaction Types

Compound B is 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 E . . . radical C

radical C . . . radical E

radical B . . . radical E

radical D . . . radical A

radical D . . . radical E

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 #1 : Carbocations

What intermediate is involved in the conversion of compound B to compound C?

Mcat_1

Possible Answers:

Tertiary carbocation

Secondary carbocation

Secondary radical

Tertiary carbanion

Tertiary radical

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 #1421 : Mcat Biological Sciences

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 low temperatures

No, never

Yes, only at high temperatures

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.

Because there are nine moles of reactant and eleven 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 #11 : Reaction Types

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

Possible Answers:

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

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

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

The heat of combustion per  in cyclohexane is greater than the heat of combustion per  in cyclobutane

Correct answer:

The heat of combustion per  in cyclohexane is greater than the heat of combustion per  in 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 per  group in the ring. Even though cyclohexane has more carbons than cyclobutane, the heat of combustion per  group in cyclobutane will be greater compared to cyclohexane due to cyclobutane's ring strain. 

Example Question #11 : Reaction Types

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:

The compounds will have the same heat of reduction

More information is required

Benzene

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 #11 : Reactions

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:

Rank these four alkenes in terms of their stability, with the least stable compound on the left, and the most stable compound on the right.

Possible Answers:

A < B < C < D

D < C < B < A

A < D < B < C

D < B < C < A

A < C < B < D

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.

Since the products are the same for each alkene, any 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 #1422 : Mcat Biological Sciences

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

Possible Answers:

Halogenation

Initiation

Termination

Propagation

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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