Answer:
Models help scientists understand processes in the natural world by providing simpler representations of those processes.
Answer:
The reaction rate of the both questions remain unchanged.
Explanation:
For question 1: The reaction 1-iodo -2- methylbutane with cyanide ion is an SN2 reaction because the Alkyl halide is a primary alkyl halide. The rate of reaction is dependent on concentration of the nucleophile and the alkyl halide at the same. For the rate of reaction to be affected (increased or decreased), the concentration of nucleophile and the alkyl halide have to be altered.
For question 2: The reaction of 2-iodo -2- methylbutane with ethanol is an SN1 reaction because the Alkyl halide is a tertiary alkyl halide. There are two-step reaction mechanism in this reaction. The first step is the rate determining step which determines the extent of the reaction and hence the rate of reaction. For the rate of reaction to be affected (increased or decreased), the concentration of the Alkyl halide alone will be altered. The rate of reaction is independent of the concentration of the nucleophile.
Answer:
Isopropyl propionate
Explanation:
1. Information from formula
The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.
2. Information from the spectrum
(a) Triplet-quartet
A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group
(b) Septet-doublet
A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group
(c) The rest of the molecule
The ethyl and isopropyl groups together add up to C₇H₁₂.
The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.
The compound is either
CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.
(d) Well, which is it?
The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.
The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.
The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.
We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.
3. Summary
My peak assignments are shown in the diagram below.
Answer:
1. It will most likely die
2. The ostrilope with level 10 armor will most likely reproduce more offspring because it is adapted to the environment.