Using the data in Table III of Problem ref 11.2, calculate the expected product composition from

Using the data in Table III of Problem ref 11.2, calculate...

Key Concepts

Minimization of Polyhalogenation

This concept involves the use of excess hydrocarbon to limit further substitution reactions once a single halogen has been introduced. By keeping the concentration of the hydrocarbon high relative to the halogen, the reaction conditions favor monohalogenation over polyhalogenation, thereby simplifying the product mixture and making the analysis of regioselectivity more straightforward.

Hydrogen Abstraction and C–H Bond Reactivity

This principle focuses on the differing reactivity of various types of C–H bonds (primary, secondary, tertiary) within a molecule. The bond dissociation energies and the resulting radical stabilities determine the rate at which hydrogen atoms are abstracted during the reaction, which in turn influences the product composition in mono-halogenated products.

Free Radical Halogenation

This concept covers the mechanism by which alkanes react with halogens under the influence of light or heat, generating free radicals that propagate the reaction through a chain mechanism. It involves initiation (formation of halogen radicals), propagation (hydrogen abstraction and formation of alkyl radicals), and termination steps, and provides the framework for understanding how halogen atoms are incorporated into hydrocarbons.

Reaction Selectivity and Kinetics

This concept explains how the rate of halogenation at various positions in an alkane is influenced by both statistical factors and the inherent reactivity of different sites. The selectivity of a halogen (such as chlorine versus bromine) is critical, as bromination is more selective due to its slower reaction rate and greater sensitivity to the stability of intermediates, while chlorination is more influenced by the statistical distribution of hydrogen atoms.

Transcript

00:01 This is the answer to chapter 15 problem number 78 from the smith organic chemistry textbook.

00:09 And this problem is asking us to consider the chlorination of propane, the free radical chlorination of propane.

00:19 And we're given a lot of information and asked to draw what the product that we have is.

00:26 And so the first thing that i did was just draw propane, draw the reaction conditions, and then rewrite some of what we've been given.

00:35 So the formula of our final product is c3h6, cl2, and we get several pieces of information from that formula.

00:46 Namely, we are now able to definitively say that we have six hydrogens and two chlorines.

00:52 So we know that we've added two chlorines to this molecule and lost two hydrogens from our starting material.

01:02 And so the way that i started was i looked at the integrations on the nmr that were given.

01:11 And there are two signals on the nmr, one with an integration of 57 units and one with an integration of 29 units.

01:20 And so you add them together.

01:23 That adds up to 86.

01:24 We know that we have to be accounting for six hydrogens because, again, that's given to us in our formula up at the top here.

01:34 And so 86 divided by six gives you roughly 14 units per hydrogen.

01:44 And so then you look at each signal.

01:51 So 57, the signal that integrates to 57, the triplet, divided by 14 is roughly four hydrogens.

02:00 And then the quintet, with a quintet with a integration of 29.

02:04 29 divided by 14 is roughly two hydrogens.

02:07 And so the triplet accounts for four hydrogens.

02:10 The quintet accounts for two hydrogens.

02:13 The third thing to look at is the splitting patterns of these two signals.

02:19 So if you remember the n plus one rule, so n plus one tells us that if a proton has n neighbors, n neighboring protons, it's going to give us n plus one.

02:32 Plus one splitting in the signal.

02:36 So for example, we look at our triplet and we can say there are three signals in this triplet, and so it has two neighbors.

02:45 Two would be n, two plus one is three, and it's a triplet.

02:49 So we know that it has two neighbors.

02:51 It's split by two neighboring protons...

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