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Chapter 18: Q66P (page 971)

Solving the following road-map problem depends on determining the structure of A, the key intermediate. Give structures for compounds A through K.

Short Answer

Expert verified

The identity of A is hept-2-one.

Step by step solution

01

Determining the structure of A, the key intermediate

The identity of A is hept-2-one.

02

Structure of compounds from A through K

(a) hept-2-one reacts with phenylmagnesiumbromide and H3O+ forms 2-hydroxy-2phenylheptane.

hept-2-one 2-hydroxy-2phenylheptane

(b) hept-2-one reacts with a Grignard reagent and H3O+ to form another product as 3-(1-hydroxy-1'methylhexyl)cylcohexanone.

hept-2-one 3-(1-hydroxy-1'methylhexyl)cylcohexanone

(c) hept-2-one reacts with HCN to form 2-hydroxy-2-methylhetanenitirile which undergoes hydrolysis to give 2-hydroxy-2-methylhetanoicacid.

(d) pentylmagensiumbromide on treatment with ethanal and H3O+ forms 2-hydroxyheptane which on further treatment with NaOCL/HOAc undergoes oxidation to give hept-2-one.

(e) hexanoic on treatment with SOCl2 to form hexanoyl which further reacts with Gilman reagent to give hept-2-one.


(f) Hydration of hept-1-yne takes place in presence of catalyst HgSO4 , H2 SO4 / H2O to hept-2-one.

(g) 2-methyl-heptan-1-ene undergoes ozonolysis which converts the alkene into carbonyl compound ( hept-2-one).

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Most popular questions from this chapter

Which of the following compounds would give a positive Tollens test? (Remember that the Tollens test involves mild basic aqueous conditions)

(a) CH3CH2COCH3

(b) CH3CH2CHO

(c) CH3CH2CH=CHCH=CHOH

(d) CH3CH2 CH2CH(OH)OCH3

(e) CH3CH2 CH2CH(OCH3)2

(f)

Question. The mass spectrum of unknown compound A shows a molecular ion at m/z 116 and prominent peaks at m/z 87 and m/z 101. Its UV spectrum shows no maximum above 200 nm. The IR and NMR spectra of A follow. When A is washed with dilute aqueous acid, extracted into dichloromethane , and the solvent evaporated, it gives a product B. B shows a strong carbonyl signal at 1715 cm-1in the IR spectrum and a weak maximum at 274nm(E =16) in the UV spectrum. The mass spectrum of B shows a molecular ion of m/z 72. Determine the structures of A and B, and show the fragmentation to account for the peaks at m/z 87 and 101.

PROBLEM 18-5

Oxidation of cholesterol converts the alcohol to a ketone. Under acidic or basic oxidationconditions, the C=C double bond migrates to the more stable, conjugated position. BeforeIR and NMR spectroscopy, chemists watched the UV spectrum of the reaction mixture tofollow the oxidation. Describe how the UV spectrum of the conjugated product, cholest-4-en-3-one, differs from that of cholesterol.

One of these reacts with dilute aqueous acid and the other does not. Give a mechanism for the one that reacts and show why this mechanism does not work for the other one.

Predict the major products of the following reactions.

(a) (b)

(c) (d)

See all solutions

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