An ancohol A when heated with copper gives a product B not having oxygen atom. B on ozonolysis gives two isomeric products C and D.C on oxidation give a monobasic acid E, silver salt of which contain 59.6 % Ag. The structure of A is

Options

A. `{:(CH_(3)-CH-CH-CH_(2)-CH_(3)),(" | |"),(" "CH_(3)" "OH):}`
B. `{:(" "OH),(" |"),(CH_(3)-C-CH_(2)-CH_(2)-CH_(3)),(" |"),(" "CH_(3)):}`
C. `{:(CH_(3)-CH-CH_(2)-CH_(2)-CH_(2)OH),(" |"),(" "CH_(3)):}`
D. `{:(CH_(3)-CH-CH_(2)-O-CH_(2)-CH_(2)OH),(" |"),(" "CH_(3)):}`

Answer

B

Explanation

Molecular mass of silver salt of `E = (108)/(59.6)xx10=181`
RCOOAg = 181
`R = 181-(108 +12+32)=181 -152 =29`
Hence, acid salt is `CH_(3)-CH_(2)-COOAg`
Corresponding to this salt, acid is `CH_(3)-CH_(2)-COOH`, which is obtained by oxidation of (C), an ozonolysis product. Thus, (C) is `CH_(3)-CH_(2)-CH=0`
Since, (C) and (D) are isomers, D is `CH_(3)-overset(overset(CH_(3))(|))(C)=O`. On basis of C and D, we can get product B and finally A
`CH_(3)-underset(underset(underset(A)(CH_(3)))|) overset(overset(OH)(|))(C)-CH_(2)-CH_(2)-CH_(3)overset(Cu//Delta)(rarr) CH_(3)-underset(B)overset(overset(CH_(3))(|))(C)=CH-CH_(2)-CH_(3)`