Compound 𝐗 is optically inactive and has the formula C16 H16 Br2. On treatment with strong base,

Compound 𝐗 is optically inactive and has the formula C16 H16...

Key Concepts

Structural Elucidation via Reaction Sequences

Structural elucidation through reaction sequences involves using a series of chemical transformations—such as elimination, hydrogenation, and ozonolysis—to dissect an unknown compound into identifiable fragments. Analysis of changes in molecular formula and the identification of characteristic functional groups allows chemists to reconstruct the original molecular framework and understand its stereochemistry.

Ozonolysis

Ozonolysis is a reaction in which ozone cleaves carbon–carbon double bonds, breaking them into smaller fragments that often contain carbonyl groups (such as aldehydes and ketones). The nature, number, and molecular formulas of the resulting fragments are valuable clues for deducing the original structure of the alkene.

Hydrogenation Reactions

Hydrogenation involves the addition of hydrogen across carbon–carbon multiple bonds, typically performed in the presence of a catalyst such as palladium. This reaction is used to reduce unsaturated compounds like alkenes to saturated ones, and the extent of hydrogen uptake can provide information about the number and types of multiple bonds present in the molecule.

Elimination Reactions (Dehydrohalogenation)

Elimination reactions involve the removal of atoms or groups from a molecule to form multiple bonds, often requiring a strong base. In dehydrohalogenation, a base abstracts a proton adjacent to a halogen, leading to the simultaneous loss of the halide ion and the formation of a double bond, which is a common method for transforming vicinal dihalides into alkenes.

Stereochemistry and Optical Inactivity

Stereochemistry is the study of the spatial arrangement of atoms in molecules, and optical inactivity indicates that a compound does not rotate plane?polarized light. This can occur even when chiral centers are present if the molecule is arranged in a way that makes it internally symmetrical (a meso compound) or if a racemic mixture is formed.

Transcript

00:01 So we're given the following information here for three different compounds, x, y, and z.

00:07 And we want to try to figure out structures for each of these compounds with the following information.

00:13 So we see this is a pretty difficult question, but it does require us to kind of analyze the pieces of information that we're given one at a time.

00:26 So with compound x, we know that it is optically inert.

00:30 Optically inert means typically means that we're going to have a plane of symmetry in our compound.

00:38 And this is indicative of a meso compound.

00:43 So our compound is going to be achyrol, although it's likely that it's going to have chiral centers.

00:50 And so we're also given the molecular formula.

00:53 We have c -16h -16 br2.

00:56 So the fact that we have our c -16 h -16, it's going to suggest that we're going to probably have aromatic.

01:03 Hydrocarbons so it's likely that since we have our meso compound i'm going to say we're going to have two phenyl rings likely right and now we have our br2 and so this is going to suggest if we have two phenyl rings we're going to have a four carbon chain a four carbon parent chain and therefore so our br2 in order for it to be a meso compound and optically inert it's either got to be on, so, carbon 1 and 4, or it's got to be on carbon 2 or 3 in order to fulfill this optically inert requirement.

01:56 So why don't we go ahead and start drawing that now? all right, so given that information, we can draw it in the following way.

02:08 So we know we're going to have our two final rings here.

02:17 I'm going to draw in a certain way just so we can really see here the fact that we have a planar cemetery, and therefore our compound is going to be meso here.

02:39 Okay.

02:39 So now what we can do here, like i said earlier, we're either going to have our bromides on the 1 and 4 carbon here.

02:46 So that'll look like this.

03:01 Okay.

03:03 And this is very easy to see our plan of symmetry here.

03:07 So we have our plane of symmetry right here.

03:10 And so therefore we're going to have an optically inert compound that fulfills the required molecular formula.

03:17 We can also have another, we can draw this in another way as well, right? so with our bromides on the carbon number two.

03:24 So what is that going to look like? right, so we're going to get this here.

03:52 And now our bromides and hydrogens.

04:01 Go ahead and draw those in here.

04:05 And now we see, once again, we have our plan of symmetry.

04:16 And our compound is going to be optically inert.

04:19 So how do we know which one we have? well, we look at the following piece of information that we're given, right? so when we treat it with a strong base, we're going to get compound y, which is a c -16 -h -14 hydrocarbon.

04:31 So we know that since we're getting rid of two hydrogens and two bromides, we must have had two eliminations that occurred, right? so from here, i'm going to say two elm reactions.

04:50 Occurred okay and so we know that c16 h14 is going to be our only product right but if we look at this here so since we have our bromide in the 1 and 4 position here that we are going to have two elimination reactions that occur and because the only hydrogens that we can that we have as our beta hydrogens are going to be this one for the left bromidegions and this one for the right bromide.

05:27 Therefore, this is likely going to be our desired compound x.

05:33 But if we look at the right compound here, there's a lot of competing elimination reactions that can occur, right? because we have, you know, we have this hydrogen, which can react with the left bromide.

05:46 We also have this hydrogen, which could react with the left bromide.

05:49 Or, you know, for the right bromide, we could have this hydrogen or we could have this hydrogen.

05:53 So we're going to have a lot of competing elimination reactions...

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