Answer:
The different structures are shown in the attachment.
I and II - structural isomers
I and III - Structural isomers
I and IV - structural isomers
II and III - structural isomers
II and IV - structural isomers
III and IV - stereoisomers
Explanation:
The knowledge of Isomerism is tested here; there are two types of isomerism ; structural and stereoisomerism.
- Structural Isomers have similar molecular and different double bond positioning, these occurs mostly in ALKENE FAMILY.
- Stereo-isomers have the same molecular formular and similar patterns but differ in their spatial arrangement. trans and cis are typical examples of stereo-isomers.
From the question; Relationship between I and II is that they are structural isomers since they have the same molecular formula, but different bond atom arrangement and infact they are the same compound.
- Relationship between I and III is that they are structural isomers with similar molecular formular but differ in the double bond position.
- Relationship between I and IV is that they are structural isomers with similar molecular formula but different double bond arrangement.
- Relationship between II and III is that they are structural isomers with similar molecular formular but different double bond position
- Relationship between II and IV is that they are also structural isomers with the same molecular formular but different double bond position.
- Relationship between III and IV is that they are stereo-isomers with same molecular formula but different spatial arrangement, hence cis and trans.
You know 1 tbsp is 15 milliliters so you would multiply 15 and 3 together. 15 x 3 =45
So there are 45 milliliters in 3 tbsp.
Answer:
A) CH3CH2CH2CH2CH2CH2OH
Explanation:
For this question, we have the following answer options:
A) CH3CH2CH2CH2CH2CH2OH
B) (CH3CH2)2CH(OH)CH2CH3
C) (CH3CH2)2CHOHCH3
D) (CH3CH2)3COH
E) (CH3CH2)2C(CH3)OH
We have to remember the<u> reaction mechanism</u> of the substitution reaction with
. <em>The idea is to generate a better leaving group in order to add a "Br" atom.</em>
The
attacks the "OH" generation new a bond to P (O-P bonds are very strong), due to this new bond we will have a better leaving group that can remove the oxygen an allow the attack of the Br atom to generating a new C-Br bond. This is made by an <u>Sn2 reaction</u>. Therefore we will have a faster reaction with <u>primary substrates</u>. In this case, the only primary substrate is molecule A. So, <em>"CH3CH2CH2CH2CH2CH2OH"</em> will react faster.
See figure 1
I hope it helps!