Spread Spectrum (SS) encoding: a) Name four spread spectrum techniques. b) Which one of them was originally devised against eave
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This problem is asking for an explanation of what we need to know about double and triple bonds to successfully predict molecular geometries in molecules. At the end, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.
<h3>Molecular geometry:</h3>In chemistry, molecules are not necessarily flat arrangements of atoms, yet they have specific bond angles, orientations and shapes, which define the molecular geometry. In such a way, we can use the VSEPR theory in order to know the molecular geometry of a molecule; however, we first need its Lewis structure or at least the number and type of bonds to do so.
Consider water and carbon dioxide; the former has two hydrogen to oxygen bonds (O-H) and 2 lone pairs because O has six valence electrons but just 2 are bonded to complete the octet, so 4 unpaired electrons lead to two lone pairs. On the other hand, the latter has two double bonds (C=O) and 0 lone pairs because carbon has four valence electrons and they are all bonded to complete the octet.
In such a way, one can see how the double bond affected the bonding in CO2 in contrast to the H2O; situation that also applies to triple bonds, because CO2 has a linear molecular geometry whereas water has a bent one (see attached picture)
Hence, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.
Learn more about molecular geometry: brainly.com/question/7558603
Learn more about the VSEPR theory: brainly.com/question/14225705
<h3><u>Full Question:</u></h3>
The following compound has been found effective in treating pain and inflammation (J. Med. Chem. 2007, 4222). Which sequence correctly ranks each carbonyl group in order of increasing reactivity toward nucleophilic addition?
A) 1 < 2 < 3
B) 2 < 3 < 1
C) 3 < 1 < 2
D) 1 < 3 < 2
<h3><u>Answer: </u></h3>The rate of nucleophilic attack of carbonyl compounds is 2<3 <1.
Option B
<h3><u>Explanation. </u></h3>Nucleophilic attack is explained as the attack of an electron rich radical to a carbonyl compound like aldehyde or a ketone. A nucleophile has a high electron density, so it searches for a electropositive atom where it can donate a portion of its electron density and become stable.
A carbonyl compound is a hybridized carbon atom with a double bonded oxygen atom in it. The oxygen atom pulls a huge portion of electron density from carbon being very electropositive.
In a ketone, there are two factors that make it less likely to undergo a nucleophilic attack than aldehyde. Firstly, the steric hindrance of two carbon groups being attached with the carbonyl carbon makes it harder for the nucleophile to approach. Secondly, the electron push by the carbon groups attached makes the carbonyl carbon a bit less electropositive than the aldehyde one. So aldehydes are more reactive towards a nucleophilic addition reaction.