Choose the correct answer among the alternatives given,
One mole of ethyl acetate on treatment with an excess of $L i A l H_{4}$ in dry ether and subsequent acidification produces:

1 m o l

acetic acid

+ 1 m o l

ethyl alcohol.

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1 m o l

ethyl alcohol

+ 1 m o l

methyl alcohol.

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2

moles of ethyl alcohol.

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1 m o l

2 -

butanol.

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Solution

The correct option is C $2$ moles of ethyl alcohol.
$C H_{3} C O O C_{2} H_{5} E t h y l a c e t a t e L i A l H_{4} / H_{3} O^{+} - -------- \to d r y e t h e r C H_{3} C H_{2} O H + C_{2} H_{5} O H$

Ethyl Acetate on treatment with $L i A l H_{4}$ gives 2 moles of Ethyl Alcohol.

Hence,option C is correct.

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Similar questions

$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l); K_{c} = 4$

In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little conc. $H_{2} S O_{4}$ . On equilibrium being attained

1 mole of methyl alcohol is mixed with 4 moles of ethyl alcohol. The mole fraction of ethyl alcohol is:

The rate of esterification of acetic acid with methyl alcohol

(I) methyl alcohol

(II) ethyl alcohol

(III) isopropyl alcohol

(IV) tert-butyl alcohol

Q. Treatment of ethyl alcohol with Conc.

H_{2} S O_{4}

at 443 K gives

Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is 0. 25 hence the percentage of ethyl alcohol by weight is:

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Choose the correct answer among the alternatives given,One mole of ethyl acetate on treatment with an excess of LiAlH4 in dry ether and subsequent acidification produces:

The correct option is C 2 moles of ethyl alcohol.CH3COOC2H5EthylacetateLiAlH4/H3O+−−−−−−−−−→dryetherCH3CH2OH+C2H5OHEthyl Acetate on treatment with LiAlH4 gives 2 moles of Ethyl Alcohol.Hence,option C is correct.

Choose the correct answer among the alternatives given,
One mole of ethyl acetate on treatment with an excess of $L i A l H_{4}$ in dry ether and subsequent acidification produces: