Organic Chemistry Questions and Answers

Answer: B

The most acidic hydrogen is in ethyne, HC≡CH.

C-H acidity depends on how well the carbon holds the lone pair left behind. An sp carbon has 50% s character, so its lone pair sits close to the nucleus and the acetylide anion is relatively stable; ethyne has a pK_a of about 25.

The C-H bonds in ethene and allene are on sp² carbons (33% s character), and those in ethane on sp³ carbons (25%), so their conjugate bases are far less stable; ethene and ethane have pK_a values of roughly 44 and 50. Allene's central carbon is sp, but it carries no hydrogen, so it does not help.

Answer: E

There are 14 and 1 sigma and pi bonds.

3-Methylbut-1-ene is \(CH_{2}=CH-CH(CH_{3})-CH_{3}\), formula \(C_{5}H_{10}\).

\[ \begin{aligned} \sigma(\text{C-H}) &= 10 \\ \sigma(\text{C-C}) &= 4 \\ \text{total } \sigma &= 14,\quad \pi = 1 \end{aligned} \]

The single C=C double bond is one sigma plus one pi bond (its sigma is counted among the 4 C-C bonds). Options with 2 pi bonds would need a diene or alkyne, and 1 and 14 swaps the numbers.

Answer: C

The hybridisations are sp³, sp², sp², sp.

Count the electron domains on each carbon in CH₃–CH=CH–C≡N:

1. CH₃: four single bonds, sp³.
2. CH=: one double bond, sp².
3. =CH–: the other end of the double bond, sp².
4. C≡N: a triple bond, sp.

Options with sp on the second or third carbon treat a double-bonded carbon as triple-bonded, and option D gives the nitrile carbon sp² instead of sp.

Answer: D

The group showing the –R effect is –CN.

A group shows a negative resonance (–R or –M) effect when it withdraws electrons from a conjugated system through a multiple bond to a more electronegative atom. The C≡N group pulls π-electron density towards nitrogen, like –NO₂, –CHO and –COOH.

–OH, –OR, –NH₂ and –OCOR all have a lone pair on the atom attached to the ring, which they donate into the conjugated system: they show the +R effect.

Answer: D

The group showing a -R effect is \(\mathrm{-NO_{2}}\).

The resonance (mesomeric) effect is -R when a group withdraws pi electrons from the conjugated system it is attached to. The nitro group has a N=O double bond conjugated with the ring and pulls electron density towards its oxygen atoms, so it is a -R group (as are -CHO, -COOH and -CN).

\(\mathrm{-OH}\), \(\mathrm{-OCH_{3}}\), \(\mathrm{-NH_{2}}\) and \(\mathrm{-NHCOCH_{3}}\) all have a lone pair on the atom joined to the ring, which they donate into the pi system. That is a +R effect, which is why these groups activate a benzene ring towards electrophilic substitution, while the nitro group deactivates it.

Answer: C

The percentage of carbon is 25%.

Carbon is 12 parts in 44 of CO₂, so the mass of carbon in 0.11 g of CO₂ is

\[ \begin{aligned} m_{C} &= \frac{12}{44} \times 0.11 = 0.03\ \text{g} \\ \%C &= \frac{0.03}{0.12} \times 100 = 25\% \end{aligned} \]

Dividing 0.11 by 0.12 without converting to carbon would give over 90%, and the other values are arithmetic slips around the correct factor of \(\frac{12}{44}\).

Answer: E

The compound is cyclobutane.

The empirical formula mass of CH₂ is 14, so the multiplier is \(\frac{56}{14} = 4\) and the molecular formula is C₄H₈. Of the options, only cyclobutane has this formula.

n-Butane and 2-methylpropane are C₄H₁₀ (58), propene is C₃H₆ (42, which has the right empirical formula but the wrong mass), and propane is C₃H₈.

Answer: A

The H-C-H bond angle in ethene is 117.6°.

Each carbon in ethene is sp² hybridised, so the ideal angle is 120°. The C=C double bond has more electron density than a C-H bond and repels the C-H bonds more strongly, squeezing the H-C-H angle slightly below 120°, to 117.6° (with H-C-C about 121°).

121° is the H-C-C angle, not H-C-H. 110° and 105° are near the sp³ and water-like values, and 119° has no basis in the measured structure.

Answer: E

The percentage of carbon is 25%.

All the carbon ends up in \(CO_{2}\), which is \(\frac{12}{44}\) carbon by mass:

\[ \begin{aligned} m_{C} &= \frac{12}{44} \times 0.88 = 0.24\ \text{g} \\ \%C &= \frac{0.24}{0.96} \times 100 = 25\% \end{aligned} \]

The water figure is needed only for hydrogen. Taking the \(CO_{2}\) mass itself as a share of 0.96 g, or using 0.24 g out of a wrong total, gives the other options.

Answer: B

The empirical formula is CH₂O.

Take 100 g of the compound. The remainder after carbon and hydrogen, \(100 - 40 - 6.7 = 53.3\) g, is oxygen. Divide each mass by its atomic mass:

\[ \begin{aligned} \text{C}: \tfrac{40}{12} &= 3.33 \\ \text{H}: \tfrac{6.7}{1} &= 6.7 \\ \text{O}: \tfrac{53.3}{16} &= 3.33 \end{aligned} \]

Dividing by the smallest, 3.33, gives C : H : O = 1 : 2 : 1, so the formula is CH₂O (as in formaldehyde, acetic acid and glucose). Ignoring the oxygen would give CH₂, which is not offered; the 53.3% left over has to be oxygen.