In the following sequence of reactions-Z PCl 5- X Alc- KOH - Y i conc- H 2 SO 4- ii H 2 O - boil ZZ is-A- CH 33 CCH 2 OHB- CH 3 CH 2 CH 2 OHC- CH 3 CHOHCH 3D- CH 3 CH 2 CH 2 CH 2 OH

Since alkenes give Markovnikov's addition products, generally 1^∘ alcohols (a), (c) and (d) cannot be obtained by hydration. Thus, 'Z' must be 2 propanol, i.e., ...

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Standard XII

Chemistry

Sulphuric Acid

In the follow...

Question

In the following sequence of reactions,
$Z P C l_{5} - -- \to X Alc. KOH - ----- \to Y (i) conc. H_{2} S O_{4} - -------- \to (i i) H_{2} O, boil Z$
$Z$ is:

C H_{3} C H_{2} C H_{2} O H

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C H_{3} C H O H C H_{3}

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C H_{3} C H_{2} C H_{2} C H_{2} O H

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(C H_{3})_{3} C C H_{2} O H

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Solution

The correct option is B $C H_{3} C H O H C H_{3}$
Since alkenes give Markovnikov's addition products, generally $1^{\circ}$ alcohols (a), (c) and (d) cannot be obtained by hydration. Thus, $^{'} Z^{'}$ must be 2-propanol, i.e.,

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will be:

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Major product (Z) will be ?

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(C H_{3})_{2} C H C O O H (i) L i A l H_{4} ⟶ (i i) H_{2} O P B r_{3} ⟶ K C N ⟶ D M S O H_{2} O, H^{+} ⟶ H e a t

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In the following sequence of reactions, ZPCl5−−−→XAlc. KOH−−−−−−→Y(i) conc. H2SO4−−−−−−−−−→(ii) H2O, boilZ Z is:

The correct option is B CH3CHOHCH3Since alkenes give Markovnikov's addition products, generally 1∘ alcohols (a), (c) and (d) cannot be obtained by hydration. Thus, ′Z′ must be 2-propanol, i.e.,

In the following sequence of reactions,
$Z P C l_{5} - -- \to X Alc. KOH - ----- \to Y (i) conc. H_{2} S O_{4} - -------- \to (i i) H_{2} O, boil Z$
$Z$ is:

The correct option is B $C H_{3} C H O H C H_{3}$
Since alkenes give Markovnikov's addition products, generally $1^{\circ}$ alcohols (a), (c) and (d) cannot be obtained by hydration. Thus, $^{'} Z^{'}$ must be 2-propanol, i.e.,