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Chapter 11: Q64P (page 591)

Unknown Q is determined to have a molecular formula of C6H12O. Q is not optically active, and passing it through a chiral column does not separate it into enantiomers. Q does not react with Br2, nor with cold, dilute KMnO4 , nor does it take up H2under catalytic hydrogenation. Heating of Q with H2SO4 gives product R, of formula C6H10, which can be separated into enantiomers. Ozonolysis of a single enantiomer of R produces S, an acyclic, optically active ketoaldehyde of formula C6H10O2. Propose structures for compounds Q, R, and S, and show how your structures would react appropriately to give these results.

Short Answer

Expert verified

The molecular formula of the compound is C6H12O. The compound Q has one element of unsaturation. There may be a ring or C=C or C=O. There is no asymmetric carbon in Q and cannot be separated into enantiomers.

Step by step solution

01

About the given information

The molecular formula of the compound is C6H12O. The compound Q has one element of unsaturation. There may be a ring or C=C or C=O. There is no asymmetric carbon in Q and cannot be separated into enantiomers.

02

About the given information

Compound Q has no reaction with Br2 , KMnO4, H2 . Thus, Q is not an alcohol, not C=O, and C=C. Q undergoes a reaction with H2SO4and loses H2O. Thus, Q has a ring.

03

About the reaction

The compound R has a C=C, that is, it ozonolyses to give one acyclic product. The transformation from S to R has a C=C in the ring. R has enantiomers and the dehydration produces C=C creating an asymmetric carbon. S is a ketoaldehyde that gives a 4-membered ring.

Thus, the reaction is as follows:

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Most popular questions from this chapter

Predict the products formed by periodic acid cleavage of the following diols

(a)

(b)

(c)

(d)

(a) The reaction of butan-2-ol with concentrated aqueous HBr goes with partial racemization, giving more inversion than retention of configuration. Propose a mechanism that accounts for racemization with excess inversion.

(b)Under the same conditions, an optically active sample of trans-2-bromocyclopentanol reacts with concentrated aqueous HBr to give an optically inactive product, (racemic) trans-1,2-dibromocyclopentane. Propose a mechanism to show how this reaction goes with apparently complete retention of configuration, yet with racemization. (Hint: Draw out the mechanism of the reaction of cyclopentene with in water to give the starting material, trans-2- bromocyclopentanol. Consider how parts of this mechanism might be involved in the reaction with HBr.)

Predict the products of the reactions of the following compounds with:

(1) chromic acid or excess sodium hypochlorite with acetic acid.

(2)PCC or NaOCl (1 equivalent) with TEMPO.

(a)cyclohexanol (b)1-methylcyclohexanol

(c)cyclopentylmethanol (d)cyclohexanone

(e)cyclohexane (f)1-phenylpropan-1-ol

(g)hexan-1-ol (h)acetaldehyde, CH3CHO

The compound shown below has three different types of OH groups, all with different acidities. Show the structure produced after this compound is treated with different amounts of NaH followed by a methylating reagent. Add a brief explanation.

(a)1 equivalent of NaH, followed by 1 equivalent of CH3l and heat

(b)2 equivalents of NaH, followed by 2 equivalents of CH3l and heat

(c) 3 equivalents of NaH, followed by 3 equivalents of CH3l and heat

Two products are observed in the following reaction.

(a) Suggest a mechanism to explain how these two products are formed.

(b) Your mechanism for part (a) should be different from the usual mechanism of the reaction of SOCl2 with alcohols. Explain why the reaction follows a different mechanism in

this case.
See all solutions

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