A hydrocarbon A, C9 H10, is treated with N -bromosuccinimide...

A hydrocarbon $A, \mathrm{C}_{9} \mathrm{H}_{10},$ is treated with $N$ -bromosuccinimide to give a single monobromo compound $B$. When $B$ is dissolved in aqueous acetone it reacts to give two nonisomeric compounds: $C$ and $D .$ Catalytic hydrogenation of $D$ gives back $A,$ and $C$ can be separated into enantiomers. When optically active $C$ is oxidized with $\mathrm{CrO}_{3}$ and pyridine, an optically inactive ketone $E$ is obtained. Vigorous oxidation of $A$ with $\mathrm{KMnO}_{4}$ affords phthalic acid (structure in Problem 17.16, p. 858). Propose structures for compounds $A$ through $E,$ and explain your reasoning.

Key Concepts

Oxidation of Benzylic Positions Leading to Carboxylic Acids

Vigorous oxidation of benzylic positions—particularly in polycyclic or fused ring systems where the benzylic carbon is part of a larger structural unit—leads to the formation of carboxylic acid groups. In aromatic compounds with alkyl substituents at positions that allow for the formation of di?carboxylated products upon oxidation, this process is exemplified by the conversion to compounds such as phthalic acid. This highlights the unique reactivity of benzylic hydrogens in harsher oxidative conditions.

Stereochemistry in Cyclic Systems

When a substitution reaction at a benzylic position generates a new chiral center, the product can exist as a pair of enantiomers. The separation and subsequent reactivity (such as oxidation to an achiral ketone) of these enantiomers underscore the importance of stereochemistry in cyclic and fused ring systems. The study of how the creation or loss of chirality affects the physical and chemical properties is a central theme in organic synthesis and reaction mechanism analysis.

Oxidation of Secondary Alcohols to Ketones

Secondary alcohols can be oxidized to ketones using reagents such as chromium trioxide in the presence of pyridine. In this oxidation, the chiral center present in the secondary alcohol becomes a planar carbonyl group in the ketone, leading to a loss of optical activity. This transformation is commonly used to probe the stereochemical integrity of a chiral center and to change its reactivity.

Benzylic Bromination

This concept involves the selective replacement of a benzylic hydrogen (a hydrogen on a carbon that is attached to an aromatic ring) with a bromine atom using N?bromosuccinimide (NBS). Because the benzylic position is activated by resonance with the aromatic ring, it is much more reactive than other C–H bonds, so the reaction typically proceeds with high selectivity and without brominating the aromatic ring itself.

Solvolysis and Competing Reaction Pathways of Benzylic Halides

Once a benzylic bromide is formed, it can undergo different reaction pathways when treated with nucleophilic solvents. Under solvolytic conditions, the benzylic halide may react via substitution to yield a product bearing a new functional group (often an alcohol if water is present), or it may undergo elimination to yield an alkene. These competing pathways can lead to products that differ in connectivity and stereochemistry.

Catalytic Hydrogenation of Unsaturated Compounds

Catalytic hydrogenation is a process in which unsaturated bonds, such as those in alkenes, are reduced to saturated bonds using hydrogen gas in the presence of a catalyst. In the context of competing reaction pathways, hydrogenation can be used to selectively reduce an unsaturated product (for example, an alkene) back to its saturated form, thereby ‘repairing’ a bond that was created in an elimination reaction.

Transcript

00:01 Hello students, welcome back.

00:02 Here we are given hydrocarbon opti formula c6 h10.

00:09 Okay, we are given this compound and we need to tell which compound it can be and we are given certain conditions.

00:16 Okay.

00:17 So we are given that this compound reacts with sodiumide and the same on ozonolases followed by hydrogen peroxide oxidation gives two molecule of carboxylic acids.

00:29 One, being optically active okay so for that first of all we see it's an alkyne right so uh we are given uh five options okay first of all these options we'll check one by one which can best suit it so our first option is one hexine so let's say this is one two six three four and five and six these are six carbon and one hexine will be this right if we look at the second um i could look at the second option, it is given 2 hexine.

01:05 So if i draw it, it will be something like this, okay? two hexine and next is 3 hexine.

01:12 So 3 hexine will be something like this.

01:15 And then we are given 3 -3 -dyl -1 butyme.

01:19 Okay, so here we are given butyne.

01:23 So i'll draw 1, 2, 3 -4 and 3 -3 -dymethyl butyne.

01:29 So there is a methyl on the third carbon.

01:34 So there will be one methyl and then the another methyl and this is the iron group.

01:41 And the last compound which we are given is three methyl, one pentine.

01:48 Okay...