Classify the following alcohol as primary,

Chemistry

1 answer:

Nata [24]2 years ago

3 0

The classification of the alcohols gives;

Compound 1 - Primary alcohol
Compound 2 - Tertiary alcohol
Compound 3 - Secondary alcohol
Compound 4 - Secondary alcohol

<h3>What are alcohols?</h3>

Organic compounds occurs in families. The family of compounds is called a homologous series. The homologous series always have a functional group. The functional group is the atom, group of atoms or bond that is responsible for the chemical reactivity of the members of a given homologous series.

Now we know that the alcohols are those organic compounds that contains the -OH group. The could be aliphatic or alicyclic compounds. We shall now proceed to name the kind of alcohols that each of the compounds shown are;

Compound 1 - Primary alcohol
Compound 2 - Tertiary alcohol
Compound 3 - Secondary alcohol
Compound 4 - Secondary alcohol

Learn ore about alcohols:brainly.com/question/4698220

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Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .