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3 years ago

Why can't 1−methylcyclohexanol be prepared from a carbonyl compound by reduction? select the single best answer?

Chemistry

1 answer:

forsale [732]3 years ago

4 0

1−methylcyclohexanol is a tertiary alcohol. Tertiary Alcohols are synthesized by either reacting Ketone with Organometallic compounds like Grignard reagent or by hydration of substituted alkenes. 1−methylcyclohexanol can not be synthesized by reduction of carbonyl compound because it is not possible to have a starting carbonyl compound having carbonyl group along with three other alkyl groups (as carbon can only form 4 bonds).

Result:
Tertiary alcohols don't contain a hydrogen atom at carbon attached to hydroxyl group that is why it is not possible to synthesize 1−methylcyclohexanol by reduction of carbonyl compound.

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Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

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$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

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Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

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Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

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