Sketch out the n.m.r. spectra, including spin-spin splitting (if any), expected for the following compounds: (a) \(\mathrm{BrCH}=\mathrm{C}=\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}=\mathrm{C}=\mathrm{CHCH}_{2} \mathrm{OCH}_{3}\)

: For the first compound, \(\mathrm{BrCH}=\mathrm{C}=\mathrm{CH}_{2}\), we have three distinct types of hydrogen atoms: 1. The two hydrogen atoms connected to the terminal double-bonded carbon (H_a): They are equivalent and form one signal. 2. The single hydrogen atom connected to the carbon bonded with bromine and double-bonded to carbon (H_b): This forms another signal. 3. No neighboring hydrogen atoms for H_b, so no splitting. H_a has one neighboring hydrogen atom, H_b, so it will show a doublet due to spin-spin coupling.

: For the first compound, \(\mathrm{BrCH}=\mathrm{C}=\mathrm{CH}_{2}\): 1. We have two signals: one for the two equivalent H_a atoms, and one for the single H_b atom. 2. H_a will have high-field signal due to the electron density from the double bond, with a doublet splitting pattern (n+1, where n is the number of neighboring hydrogen atoms, 1 in this case). 3. H_b will have a low-field signal due to the deshielding electron-withdrawing effect of the bromine atom, with no splitting since there are no neighboring hydrogen atoms.

: For the second compound, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\), we have three distinct types of hydrogen atoms: 1. The three hydrogen atoms in the methyl group (H_a): They are equivalent and form one signal. 2. The two hydrogen atoms in the methylene group (H_b): They are equivalent and form another signal. 3. The single hydrogen atom connected to the triple-bonded carbon (H_c): This forms another signal. 4. H_a has two neighboring hydrogen atoms in H_b, so it will show a triplet splitting pattern. H_b has three neighboring hydrogen atoms in H_a and will show a quartet splitting pattern. H_c will have no splitting as it has no neighboring hydrogens.

: For the second compound, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\): 1. We have three signals: one for the three equivalent H_a atoms, one for the two equivalent H_b atoms, and one for the single H_c atom. 2. H_a will have a low-field signal due to the shielding effect of the nearby electronegative carbon triple bond, with the triplet splitting pattern. 3. H_b will have a middle-field signal, with a quartet splitting pattern. 4. H_c will have the high-field signal due to the electron density from the triple bond, with no splitting.

: For the third compound, $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}=\mathrm{C}=\mathrm{CHCH}_{2} \mathrm{OCH}_{3}$, we have five distinct types of hydrogen atoms: 1. Nine hydrogen atoms in the three methyl groups (H_a): All are equivalent and form one signal. 2. Two hydrogen atoms attached to the terminal double-bonded carbon (H_b): Equivalent and form another signal. 3. The single hydrogen atom attached to the carbon atom bonded to the three methyl groups (H_c): Forms another signal. 4. Two hydrogen atoms in the methylene group (H_d): They are equivalent and form another signal. 5. Three hydrogen atoms in the methoxy group (H_e): They are equivalent and form another signal. 6. H_a, H_e will not have any splitting. H_c will have a doublet due to its two neighboring H_b hydrogen atoms. H_b will have a doublet due to one neighboring H_c. H_d will have a triplet splitting pattern due to two neighboring H_b atoms.

: For the third compound, $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}=\mathrm{C}=\mathrm{CHCH}_{2} \mathrm{OCH}_{3}$: 1. We have five signals: one for the nine equivalent H_a atoms, one for the two equivalent H_b atoms, one for H_c, one for the two equivalent H_d atoms, and one for the three equivalent H_e atoms. 2. H_a will have a high-field signal due to the electron-donating effect of the nearby carbon double bond, with no splitting. 3. H_b will have a middle-field signal, with a doublet splitting pattern. 4. H_c will have a low-field signal due to the deshielding, electron-withdrawing effect of the three methyl groups, with a doublet splitting pattern. 5. H_d will have a middle-field signal, with a triplet splitting pattern. 6. H_e will have a middle-field signal as well, with no splitting since they have no neighboring hydrogens.