MP Board · Class 12 · Chemistry · Alcohols, Phenols and EthersA comprehensive analysis of an organic compound containing carbon, hydrogen, and oxygen gives a molar mass of $88 \text{ g/mol}$. When this ether is cleaved by hot concentrated hydroiodic acid (HI), it yields two moles of an alkyl iodide corresponding to the exact same carbon skeleton. Identify the structure of the ether, write the step-by-step reactions involved in its cleavage, and calculate the exact mass of the alkyl iodide produced when $17.6 \text{ g}$ of this ether reacts completely with excess HI.

Step 1: Identification of the Ether\nThe general formula for dialkyl ethers is $C_nH_{2n+2}O$.\nGiven molar mass = $88 \text{ g/mol}$.\nLet's calculate the value of $n$ using the general molecular mass formula:

$12n + 1(2n+2) + 16 = 88$ $12n + 2n + 2 + 16 = 88$ $14n + 18 = 88$ $14n = 70$ $n = 5$ \nSo, the molecular formula of the ether is $C_5H_{12}O$.\nSince the cleavage with hot concentrated HI yields two moles of the same alkyl iodide, it implies that the ether is symmetrical containing identical alkyl groups on both sides of the oxygen atom. An ether with 5 carbon atoms and symmetrical structure can only be di-ethyl methyl-type or similar symmetric variations. Wait, a 5-carbon symmetric ether cannot have identical halves unless it is diethoxy-type, but $C_5$ means an odd number of carbons, so let's analyze symmetric ethers: symmetric ethers are formed by identical alkyl groups, meaning $2 \times x = 5$ (not possible directly for symmetric simple ethers unless mixed, but the problem states it yields two moles of an alkyl iodide corresponding to the exact same carbon skeleton - meaning it splits into identical alkyl groups, which implies 2.5 carbons each? No, let's look at 2-ethoxypropane or di-isopropyl ether: di-isopropyl ether has 6 carbons ($C_6H_{14}O$). Let's check diethyl ether ($C_4H_{10}O$) molar mass = 74. Let's check dipropyl ether ($C_6H_{14}O$) molar mass = 102.\nLet's re-verify the molar mass $88 \text{ g/mol}$ for an ether: $C_5H_{12}O$ is pentyl ether? No, dibutyl is $C_8$. Let's check $C_5H_{12}O$: Pentanol isomers have molar mass 88. Is there an ether with molar mass 88? \nLet's check 2-ethoxypropane: $C_5H_{12}O$ -> $CH_3-CH_2-O-CH(CH_3)2$. Cleavage gives ethyl iodide and isopropyl iodide. \nWait, the question states: "yields two moles of an alkyl iodide corresponding to the exact same carbon skeleton." Let's assume the ether is symmetrical like di-isopropyl ether, but its molar mass is $6 \times 12 + 14 + 16 = 102$. \nLet's check $C_5H{12}O$ ethers: Methyl isobutyl ether, etc. If it gives two moles of an alkyl iodide, let's assume the compound is 2-methoxy-2-methylpropane (methyl tert-butyl ether, MTBE): $C_5H_{12}O$, molar mass = $5 \times 12 + 12 + 16 = 88 \text{ g/mol}$. \nCleavage of MTBE with HI: $CH_3-O-C(CH_3)_3 + HI \rightarrow CH_3OH + (CH_3)_3CI$\nWith excess HI, $CH_3OH + HI \rightarrow CH_3I + H_2O$.\nThus, it yields methyl iodide and tert-butyl iodide. Both are alkyl iodides derived from the respective alkyl groups.

Step 2: Reactions for Cleavage with HI

1. Protonation of ether: $CH_3-O-C(CH_3)_3 + HI \rightleftharpoons CH_3-O^+H-C(CH_3)_3 + I^-$
2. Nucleophilic attack by iodide ion: Since the tert-butyl cation is extremely stable, $S_N1$ mechanism predominates, forming tert-butyl iodide and methanol: $CH_3-O^+H-C(CH_3)_3 \rightarrow CH_3OH + (CH_3)_3C^+$ $(CH_3)_3C^+ + I^- \rightarrow (CH_3)_3CI$
3. Reaction of methanol with excess HI: $CH_3OH + HI \rightarrow CH_3I + H_2O$

Step 3: Mass Calculation of Alkyl Iodide\nMolecular formula of ether (MTBE) = $C_5H_{12}O$\nMolar mass of ether = $(5 \times 12) + (12 \times 1) + 16 = 88 \text{ g/mol}$.\nGiven mass of ether = $17.6 \text{ g}$.\nNumber of moles of ether = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{17.6}{88} = 0.2 \text{ moles}$.

\nFrom the stoichiometry of the overall reaction: $C_5H_{12}O + 2HI \rightarrow CH_3I + (CH_3)_3CI + H_2O$ 1 mole of ether produces 1 mole of methyl iodide ($CH_3I$) and 1 mole of tert-butyl iodide ($(CH_3)_3CI$).\nTotal moles of alkyl iodides produced = $0.2 \times 2 = 0.4 \text{ moles}$ (or $0.2 \text{ mol}$ of each).\nMolar mass of $CH_3I = 12 + (3 \times 1) + 127 = 142 \text{ g/mol}$.\nMolar mass of $(CH_3)_3CI = (4 \times 12) + (9 \times 1) + 127 = 48 + 9 + 127 = 184 \text{ g/mol}$. \nMass of $CH_3I = 0.2 \text{ mol} \times 142 \text{ g/mol} = 28.4 \text{ g}$.\nMass of $(CH_3)_3CI = 0.2 \text{ mol} \times 184 \text{ g/mol} = 36.8 \text{ g}$.\nTotal mass of alkyl iodides produced = $28.4 + 36.8 = 65.2 \text{ g}$.