Draw the organic product of each reaction and classify the product as an alcohol, symmetrical et

Draw the organic product of each reaction and classify the...

Key Concepts

Functional Groups in Organic Chemistry

Understanding functional groups is essential in organic chemistry as they define the structure, reactivity, and properties of organic molecules. In the context of the problem, knowing the characteristics of alcohols and ethers (both symmetrical and unsymmetrical) allows for proper identification and classification of products formed during chemical reactions.

Alcohols

Alcohols are organic compounds characterized by the presence of a hydroxyl (–OH) group attached to a carbon atom. They are fundamental in organic synthesis, serving as starting materials or products in various reactions, and can undergo processes such as dehydration or substitution to form other functional groups.

Ethers

Ethers are a class of organic compounds where an oxygen atom is bonded to two alkyl or aryl groups (R–O–R'). Their chemical inertness relative to alcohols makes them valuable as solvents and intermediates in organic synthesis. Recognizing the structure of ethers is key in differentiating them from other oxygen-containing compounds.

Symmetrical vs Unsymmetrical Ethers

A symmetrical ether has two identical alkyl or aryl groups attached to the oxygen atom, whereas an unsymmetrical ether has two different groups. The distinction is important in reaction outcomes, particularly in synthesis, as it influences properties like boiling point, solubility, and reactivity as well as the strategies employed when synthesizing these compounds through reactions such as acid-catalyzed condensation or the Williamson ether synthesis.

Transcript

00:01 Hello, today we will be doing problem 9 .10, and this asks us to draw the organic products of each of the following reactions and classify the product as either an alcohol, a symmetric ether, or an unsymmetric ether.

00:16 So let's look at the first problem, problem a, and we see that we have a straight chain al -cane with a good leaving group, which is one of the halides, bromine, and we have a pretty good nucleophile being.

00:30 A hydroxide anion having the anion on the oxygen atom.

00:36 So we know that obviously our nucleophile is going to be this hydroxide anion and our leaving group is going to be the bromine with the electrophile being this molecule here.

00:45 So if i were to show you guys the mechanism just so we understand how we get the products, we would see something like this.

00:53 If i show that using curly arrow mechanism with the end of my curly arrow from my source of the electrons which is going to be our nucleophile this hydroxid anion it will go in and backside attack the carbon that is directly bound to our good leaving group which is bromine and whenever you make a bond you must break a bond so i just made a bond and i need to break a bond and now that shows that we have a concerted one -step reaction suggesting that this is going to be an s &2 reaction to give us our final product, which is going to be this.

01:45 So we have our final product being this with obviously the leaving group falling off to form bromide ion.

01:53 So we have butanol with our bromide ion.

01:57 So this is going to be an alcohol because we have obviously a hydroxide group, which is also known as an alcohol.

02:04 So using that same methodology as before, we have our nucleophile.

02:09 Which is electron rich and it is formal negative charge so we will show the end of our curly arrow being from our nucleophile attacking with a double -headed arrows remember the two sides the two heads of the arrow represents the two electrons that are going to be used for attack so these two electrons are represented by each side of this head of the arrow that is attacking this carbon obviously when you make a bond you must break a bond so the chlorine will fall off giving us our final product being this ether right here and obviously our chlorine fell off so we need to draw that byproduct so what we have is an ether however the number of carbons on the left side of the ether is not the same as the number of carbons on the right side of ether so what we call this is an unsymmetrical ether and now if you look at problem c we have a cyclohexane attached to hydrocarbons with iodine, another halogen leaving group, good leaving group, with obviously a formal negatively charged oxygen acting as our nucleophile.

03:29 So this would be a hydroxide ion, which got deprotonated to give us o minus.

03:35 Same thing...

Please give Ace some feedback