ACS Organic Structure Bonding Quiz: 20 Free Questions

Correct answer: D. The lone pair on nitrogen requires more space than a bonding pair, compressing the N-H bonds.

A lone pair occupies a larger region of space than a bonding pair, and its stronger repulsion squeezes the N-H bond pairs closer together, reducing the angle from the ideal tetrahedral value. Electronegativity affects bond polarity, not the observed geometric compression.

From practice test 8 of our ACS Organic study guide (question 49). Original question, not an official exam item.

Correct answer: C. meta-xylene

Two substituents at positions 1 and 3 are meta, so the common name is meta-xylene. Ortho-xylene is 1,2-dimethylbenzene and para-xylene is 1,4-dimethylbenzene.

From practice test 7 of our ACS Organic study guide (question 15). Original question, not an official exam item.

Correct answer: B. The positive charge in the allyl cation is delocalized over two terminal carbons by resonance.

The allyl cation has two equivalent resonance contributors that place the positive charge on the two terminal carbons, so delocalization stabilizes the ion. Hyperconjugation is a secondary factor, and the allyl cation is not an aromatic six-electron system.

From practice test 3 of our ACS Organic study guide (question 13). Original question, not an official exam item.

Correct answer: B. 27.6 kcal/mol

The difference in heat of combustion is 9.2 kcal/mol per CH2 group, and cyclopropane has three CH2 groups. Multiplying 9.2 by 3 gives 27.6 kcal/mol of total ring strain.

From practice test 4 of our ACS Organic study guide (question 28). Original question, not an official exam item.

Correct answer: C. The nitrogen lone pair is delocalized into the carbonyl pi system and is less available for protonation.

The amide nitrogen's lone pair is conjugated with the carbonyl group, so it is less available to accept a proton. This resonance delocalization is the primary factor; inductive effects are secondary. The amide nitrogen does have a lone pair, so that statement is false.

From practice test 11 of our ACS Organic study guide (question 41). Original question, not an official exam item.

Correct answer: B. sp3

The oxygen atom in water has two bonding pairs and two lone pairs, giving four electron groups, which corresponds to sp3 hybridization. Sp2 would result from counting only the two bonds and ignoring the lone pairs.

From practice test 10 of our ACS Organic study guide (question 69). Original question, not an official exam item.

Correct answer: D. 166.6 kcal/mol

Cyclopropane has a heat of combustion of 166.6 kcal/mol per CH2 group, whereas an unstrained straight-chain alkane gives 157.4 kcal/mol per CH2. The extra 9.2 kcal/mol per group reflects ring strain.

From practice test 4 of our ACS Organic study guide (question 25). Original question, not an official exam item.

Correct answer: D. They are attached on opposite faces of the ring.

Trans means the two substituents are attached on opposite faces of the ring, while cis means same face. Adjacency and wedge bonds do not define trans; a trans pair can be on adjacent or nonadjacent carbons, and one or both may be axial or equatorial.

From practice test 5 of our ACS Organic study guide (question 15). Original question, not an official exam item.

Correct answer: D. Alkyl groups donate electron density through hyperconjugation, spreading and stabilizing the positive charge.

Hyperconjugation from adjacent C-H and C-C sigma bonds delocalizes the positive charge into the alkyl groups, which is the major reason tertiary carbocations are more stable than primary ones. The resonance-form statement is not generally correct because these carbocations lack adjacent pi systems for resonance.

From practice test 8 of our ACS Organic study guide (question 19). Original question, not an official exam item.

Correct answer: B. but-2-yne

Dimethylacetylene is CH3-C≡C-CH3; the longest chain has four carbons and the triple bond begins at C2, giving but-2-yne. But-1-yne would require the triple bond at the end of the chain.

From practice test 7 of our ACS Organic study guide (question 17). Original question, not an official exam item.

Correct answer: D. The negative charge is delocalized over both terminal carbons in the resonance hybrid.

The negative charge in the allyl anion is delocalized across the two terminal carbons, making both positions nucleophilic. The two Lewis structures are resonance contributors, not equilibrating isomers, so the anion is a single species with distributed charge.

From practice test 11 of our ACS Organic study guide (question 2). Original question, not an official exam item.

Correct answer: C. Four carbons are in a plane and the fifth carbon flaps above or below that plane.

The envelope conformation has four coplanar carbons and one out-of-plane carbon. This partial pucker reduces torsional strain while keeping angle strain low.

From practice test 12 of our ACS Organic study guide (question 67). Original question, not an official exam item.

Correct answer: D. The nitrogen lone pair in an amide is delocalized into the carbonyl pi system and is less available to accept a proton.

In an amide, the nitrogen lone pair participates in resonance with the carbonyl pi bond, so it is not freely available for protonation. The amide nitrogen is sp2 hybridized and does retain a lone pair, but it is greatly reduced in basicity compared with an amine.

From practice test 4 of our ACS Organic study guide (question 16). Original question, not an official exam item.

Correct answer: C. more stable because its pi electrons are delocalized across the intervening single bond

Conjugation allows continuous pi-orbital overlap across the single bond between the double bonds, which stabilizes the molecule. The fact that both isomers have two double bonds does not make them equal in energy because delocalization is missing in the isolated diene.

From practice test 8 of our ACS Organic study guide (question 35). Original question, not an official exam item.

Correct answer: B. The boat has eclipsed C-H bonds on adjacent carbons and a flagpole interaction between the two end hydrogens.

In the boat, the C-H bonds on adjacent carbons are eclipsed, and the two flagpole hydrogens at the ends repel each other. The chair avoids both interactions, which is why it is more stable.

From practice test 12 of our ACS Organic study guide (question 54). Original question, not an official exam item.

Correct answer: B. Tertiary alcohol

The carbon bearing the OH group is bonded to three other carbon atoms, which defines a tertiary alcohol. A secondary alcohol would have the OH carbon bonded to only two carbon atoms, and a primary alcohol to only one.

From practice test 6 of our ACS Organic study guide (question 15). Original question, not an official exam item.

Correct answer: D. The ketone, because ketones have higher nomenclature priority than alcohols

In IUPAC priority, ketones rank above alcohols, so the ketone is cited as the parent suffix and the alcohol is cited as a hydroxy prefix. Treating the alcohol as the principal group ignores the established functional group priority order.

From practice test 6 of our ACS Organic study guide (question 5). Original question, not an official exam item.

Correct answer: A. 60 degrees; angle strain

The three carbons of cyclopropane define an equilateral triangle with 60 degree C-C-C angles, far below the 109.5 degree tetrahedral ideal. This large deviation is angle strain, the dominant strain in cyclopropane.

From practice test 4 of our ACS Organic study guide (question 43). Original question, not an official exam item.

Correct answer: C. Linear, because the central beryllium atom has two electron groups and no lone pairs.

Beryllium in BeCl2 has two bonding groups and no lone pairs, so the electron groups adopt a linear arrangement with a 180 degree bond angle. The presence of polar Be-Cl bonds does not change the geometry.

From practice test 9 of our ACS Organic study guide (question 32). Original question, not an official exam item.

Correct answer: B. The C-C-C bond angles are 60 degrees, causing severe angle strain compared with the ideal 109.5 degrees.

The triangular structure of cyclopropane forces the C-C-C bond angles to 60 degrees, far below the ideal tetrahedral angle of 109.5 degrees. This angle strain makes the ring highly reactive and prone to opening in addition reactions. The other factors do not account for the elevated reactivity.

From practice test 11 of our ACS Organic study guide (question 36). Original question, not an official exam item.